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bài 1 :
b) (x-1/2 )2 = 0
<=> x - 1/2 = 0
<=> x = 0+ 1/2
<=> x = 1/2
c) ( x - 2 ) 2 = 1
<=> x -2 = 1
<=> x = 1 +2 = 3
d) ( 2x -1 )3 = -8
<=> ( 2x - 1) 3 = ( -2 ) 3
<=> 2x - 1 = -2
<=> 2x = -2+1 = -1
<=> x = -1/2
Bài 2 :
c) 32x-1=243
<=> 32x-1= 35
<=> 2x-1 = 5
<=> 2x = 6
<=> x = 6:2 = 3
Mk chỉ giải đc như vậy thôi
bạn thông cảm nhé !
h) \(5^x+5^{x+2}=650\)
\(\Leftrightarrow5^x+5^x.5^2=650\)
\(\Leftrightarrow5^x\left(1+25\right)=650\)
\(\Leftrightarrow5^x.26=650\)
\(\Leftrightarrow5^x=25\)
\(\Leftrightarrow x=2\)
haizzz,đăng ít thôi,chứ nhìn hoa mắt quá =.=
bây định làm j ở chỗ này vậy??? có j ib ns vs nhao chớ sao ns ở đây
a)
\((3x-7)^5=0\Rightarrow 3x-7=0\Rightarrow x=\frac{7}{3}\)
b)
\(\frac{1}{4}-(2x-1)^2=0\)
\(\Leftrightarrow (2x-1)^2=\frac{1}{4}=(\frac{1}{2})^2=(-\frac{1}{2})^2\)
\(\Rightarrow \left[\begin{matrix} 2x-1=\frac{1}{2}\\ 2x-1=\frac{-1}{2}\end{matrix}\right.\Rightarrow \Rightarrow \left[\begin{matrix} x=\frac{3}{4}\\ x=\frac{1}{4}\end{matrix}\right.\)
c)
\(\frac{1}{16}-(5-x)^3=\frac{31}{64}\)
\(\Leftrightarrow (5-x)^3=\frac{1}{16}-\frac{31}{64}=\frac{-27}{64}=(\frac{-3}{4})^3\)
\(\Leftrightarrow 5-x=\frac{-3}{4}\)
\(\Leftrightarrow x=\frac{23}{4}\)
d)
\(2x=(3,8)^3:(-3,8)^2=(3,8)^3:(3,8)^2=3,8\)
\(\Rightarrow x=3,8:2=1,9\)
e)
\((\frac{27}{64})^9.x=(\frac{-3}{4})^{32}\)
\(\Leftrightarrow [(\frac{3}{4})^3]^9.x=(\frac{3}{4})^{32}\)
\(\Leftrightarrow (\frac{3}{4})^{27}.x=(\frac{3}{4})^{32}\)
\(\Leftrightarrow x=(\frac{3}{4})^{32}:(\frac{3}{4})^{27}=(\frac{3}{4})^5\)
f)
\(5^{(x+5)(x^2-4)}=1\)
\(\Leftrightarrow (x+5)(x^2-4)=0\)
\(\Leftrightarrow \left[\begin{matrix} x+5=0\\ x^2-4=0\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x+5=0\\ x^2=4=2^2=(-2)^2\end{matrix}\right.\)
\(\Rightarrow \left[\begin{matrix} x=-5\\ x=\pm 2\end{matrix}\right.\)
g)
\((x-2,5)^2=\frac{4}{9}=(\frac{2}{3})^2=(\frac{-2}{3})^2\)
\(\Rightarrow \left[\begin{matrix} x-2,5=\frac{2}{3}\\ x-2,5=\frac{-2}{3}\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=\frac{19}{6}\\ x=\frac{11}{6}\end{matrix}\right.\)
h)
\((2x+\frac{1}{3})^3=\frac{8}{27}=(\frac{2}{3})^3\)
\(\Rightarrow 2x+\frac{1}{3}=\frac{2}{3}\Rightarrow x=\frac{1}{6}\)
a) \(3^{x+1}=243\)
\(\Leftrightarrow3^{x+1}=3^5\)
\(\Leftrightarrow x+1=5\Leftrightarrow x=4\)
b) \(\left(\frac{1}{2}\right)^{x+1}=\frac{1}{64}\)
\(\Leftrightarrow\left(\frac{1}{2}\right)^{x+1}=\left(\frac{1}{2}\right)^6\)
\(\Leftrightarrow x+1=6\Leftrightarrow x=5\)
c) \(\frac{81}{3x}=9\)
\(\Leftrightarrow3x=9\Leftrightarrow x=3\)
d) \(2^{x+1}+2^{x+2}=192\)
\(\Leftrightarrow2^x.2+2^x.4=192\)
\(\Leftrightarrow2^x.6=192\Leftrightarrow2^x=32\Leftrightarrow x=5\)
e) Ta có : \(\hept{\begin{cases}\left(x-1\right)^{2020}\ge0\\\left(y+2\right)^{2022}\ge0\end{cases}\Rightarrow\left(x-1\right)^{2020}+\left(y+2\right)^{2020}\ge0}\)
Mà \(\left(x-1\right)^{2020}+\left(y+2\right)^{2022}=0\)
\(\Rightarrow\hept{\begin{cases}\left(x-1\right)^{2020}=0\\\left(y+2\right)^{2022}=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=1\\y=-2\end{cases}}}\)
Bài giải
a, \(3^{x+1}=243\)
\(3^{x+1}=3^5\)
\(\Rightarrow\text{ }x+1=5\)
\(\Rightarrow\text{ }x=4\)
b, \(\left(\frac{1}{2}\right)^{x+1}=\frac{1}{64}\)
\(\frac{1}{2^{x+1}}=\frac{1}{2^6}\)
\(2^{x+1}=2^6\)
\(\Rightarrow\text{ }x+1=6\)
\(\Rightarrow\text{ }x=5\)
c, \(\frac{81}{3x}=9\)
\(27x=81\)
\(x=3\)
d, \(2^{x+1}+2^{x+2}=192\)
\(2^{x+1}\left(1+2\right)=192\)
\(2^{x+1}\cdot3=192\)
\(2^{x+1}=64=2^6\)
\(\Rightarrow\text{ }x+1=6\)
\(\Rightarrow\text{ }x=5\)
e, \(\left(x-1\right)^{2020}+\left(y+2\right)^{2022}=0\)
Mà \(\hept{\begin{cases}\left(x-1\right)^{2020}\ge0\\\left(y+2\right)^{2022}\ge0\end{cases}}\) với mọi x,y nên \(\hept{\begin{cases}\left(x-1\right)^{2020}=0\\\left(y+2\right)^{2022}=0\end{cases}}\Rightarrow\hept{\begin{cases}x-1=0\\y+2=0\end{cases}}\Rightarrow\hept{\begin{cases}x=1\\y=-2\end{cases}}\)
\(\Rightarrow\text{ }x=1\text{ ; }y=-2\)
mk ko chép lại đề nhé bn
b,
=>\(\left|x-\frac{1}{3}\right|+\frac{4}{5}=\left|-\frac{14}{5}\right|\)
=>\(\left|x-\frac{1}{3}\right|+\frac{4}{5}=\frac{14}{5}\) \(\Rightarrow\left|x-\frac{1}{3}\right|=2\)
\(\Rightarrow\orbr{\begin{cases}x-\frac{1}{3}=-2\\x-\frac{1}{3}=2\end{cases}\Rightarrow\orbr{\begin{cases}x=-\frac{5}{3}\\x=\frac{7}{3}\end{cases}}}\)
c,\(\Rightarrow\frac{x-1}{2013}+\frac{x-2}{2012}-\frac{x-3}{2011}-\frac{x-4}{2010}=0\)
=> \(\frac{x-1}{2013}-1+\frac{x-2}{2012}-1-\left(\frac{x-3}{2011}-1+\frac{x-4}{2010}-1\right)=0\)
=>\(\frac{x-2014}{2013}+\frac{x-2014}{2012}-\frac{x-2014}{2011}-\frac{x-2014}{2010}=0\)
=.\(\left(x-2014\right)\left(\frac{1}{2013}+\frac{1}{2012}-\frac{1}{2011}-\frac{1}{2010}\right)=0\)
Do \(\frac{1}{2013}+\frac{1}{2012}-\frac{1}{2011}-\frac{1}{2010}\ne0\)=> x-2014=0
=> x=2014
d,\(\left(x-7\right)^{x-1}-\left(x-7\right)^{x+11}=0\)
=>\(\left(x-7\right)^{x-1}.\left[1-\left(x-7\right)^{x+12}\right]=0\)
=> \(\orbr{\begin{cases}\left(x-7\right)^{x-1}=0\\1-\left(x-7\right)^{x+12}=0\end{cases}}\)
=> \(\orbr{\begin{cases}x-7=0\\\left(x-7\right)^{x+12}=0\end{cases}}\)
=>x=7 hoặc x-7=1 hoặc x+12=0
=> x=7 hoặc x=8 hoặc x=-12
Vậy x=7, x=8, x=-12
k,3x+x2=0
=> x(3+x)=0
=>\(\orbr{\begin{cases}x=0\\3+x=0\end{cases}}\)
=>\(\orbr{\begin{cases}x=0\\x=-3\end{cases}}\)
m, x2-2x-3(x-2)=0
=> x(x-2)-3(x-2)=0
=> (x-3)(x-2)=0
=>\(\orbr{\begin{cases}x-3=0\\x-2=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=3\\x=2\end{cases}}\)
*****Chúc bạn học giỏi*****
a) \(2x.\left(x-\frac{1}{7}\right)=0\)
b) \(2^x=16\)
c) \(3^{x+1}=9^x\)
d) \(2^{3x-1}=243\)
Giup mk nka
a) \(2x.\left(x-\frac{1}{7}\right)=0\)
=>2x=0 hoặc \(x-\frac{1}{7}=0\)
+)Nếu 2x=0
=>x=0
+)Nếu \(x-\frac{1}{7}=0\)
=>x=\(\frac{1}{7}\)
b)2x=16
=>2x=(-2)4 hoặc 2x=24
=>x=4
c)3x+1=9x
=>\(3^{x+1}=\left(3^2\right)^x\)
=>\(3^{x+1}=3^{2x}\)
=>x+1=2x
=>x+1=x+x
=>x=1
d)23x-1=243 (mình nghĩ chỗ này bạn nhầm đề rồi, 33x-1 chứ ko phải 23x-1 đâu, nếu là 23x-1 thì sẽ ko tìm được x thỏa mãn)
33x-1=243
=>33x-1=35
=>3x-1=5
=>3x=5=1=6
=>x=6:3=2